Question

Physics

Standing at the edge of a cliff 32.5m high, you do a ball. Later, you throw a second ball downward with an initial speed of 11m/s. Calculate the velocity if each of the balls when it hits the ground

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Best Answer

**Answer: The displacement of the ball**

**ℎ =**

**2 − **

**2**

**2**

**So, final speed of the first ball**

**1 = √2ℎ = √2 × 9.8 × 32.5 = 25.24**

**m**

**s**

**The change of it speed**

**∆1 = 25.24 − 0 = 25.34**

**m**

**s**

**The final speed of the second ball**

**2 = √**

**2 + 2ℎ = 27.53**

**m**

**s**

**The change of the second ball speed**

**∆2 = 27.53 − 11 = 16.53**

**m**

**s**

**Thus**

**∆1 > ∆2**

**Answer: ∆1 > ∆2**

**Explanation:**